Question

1. Let $\Delta ABC$ be similar to $\Delta DEF$ and their areas be, respectively $64c{m^2}$ and $121c{m^2}$ If EF = 15.4 cm find BC.

2. Diagonals of a trapezium ABCD with $AB||DC$ intersect each other at point O. If AB = 2CD, find the ratio of the areas of the triangles AOB and COD

Answer 1

Hint: Here in the first question since the triangles are similar then their sides area etc. are linked or related with one another. Hence we have to use these properties of similar triangles to get the required value of the side. In the second part of the question we have to find the ratio of area of two triangles of the trapezium, if we prove that these two triangles are similar then the ratio will be equal to the ratio of corresponding sides.

Complete step-by-step answer:
It is given that the triangle $\Delta ABC$ is a similar to $\Delta DEF$
Therefore using the properties of similar triangles which states that the areas of the two similar triangles is equal to the ratio of the squares of any two corresponding sides.

$\dfrac{{ar\left( {\Delta ABC} \right)}}{{ar\left( {\Delta DEF} \right)}} = \dfrac{{B{C^2}}}{{E{F^2}}}$
Substituting the value of the area and sides given in the question
\[
  \dfrac{{64}}{{121}} = {\left( {\dfrac{{BC}}{{15.4}}} \right)^2} \\
  \dfrac{8}{{11}} = \dfrac{{BC}}{{15.4}} \\
  BC = \dfrac{{15.4 \times 8}}{{11}} \\
  BC = 11.2cm \\
\]
Hence, the length of the side of triangle BC is 11.2cm

In the second part it is given that ABCD is a trapezium and AB is parallel to CD with AB = 2 CD
To solve we first need to find the ratio of the $\Delta AOB$ and \[\Delta COD\]
We will first prove that these triangles are similar
Now in $\Delta AOB$ and \[\Delta COD\]
$\angle AOB = \angle COD$ (Vertically opposite angles)
$\angle OAB = \angle OCD$ (Alternate angles are equal)
Therefore, the triangles $\Delta AOB$ and \[\Delta COD\] are similar
As we know that if two triangles are similar, then the ratio of square of its corresponding sides is equal to the ratio of its areas

$\dfrac{{ar\left( {\Delta AOB} \right)}}{{ar\left( {\Delta COD} \right)}} = {\left( {\dfrac{{AB}}{{CD}}} \right)^2}$
It is given that AB = 2 CD
$
  \dfrac{{ar\left( {\Delta AOB} \right)}}{{ar\left( {\Delta COD} \right)}} = {\left( {\dfrac{{2CD}}{{CD}}} \right)^2} \\
  \dfrac{{ar\left( {\Delta AOB} \right)}}{{ar\left( {\Delta COD} \right)}} = \dfrac{4}{1} \\
$
Hence, the ratio areas of the triangles is 4:1

Note- In order to solve these types of questions, you need to remember the properties of the triangles and the need to have information about congruence and similarity of the triangles with the congruence and similarity rules. In the above question, we used some properties of angles such as alternate angles, vertically opposite angles which are angles opposite to each other when two lines cross.